# The eigenfunctions of $\Delta \colon H_0^1(\Omega)\cap H^2(\Omega) \to L^2(\Omega)$ form an orthonormal basis?

Let $\Omega$ be a bounded open subset of $\mathbb{R}^n$ with smooth boundary. Consider the Dirichlet Laplace operator $$\begin{cases} D(A)=H^1_0(\Omega)\cap H^2(\Omega)\\ Au=\Delta u. \end{cases}$$ I usually see in some references that the eigenfunctions of this operator are an orthonormal basis for the Hilbert space $L^2(\Omega)$. It is stated in Wikipedia that this result follows from the spectral theorem on compact self-adjoint operators, applied to the inverse of the Laplacian (which is compact, by the Poincaré inequality and Rellich–Kondrachov theorem).

1) first why the operator $A$ is invertible

2) why the inverse is a bounded operator. (my guess is that the boundedness maybe follows from the closed graph theorem, $A$ closed operator implies $A^{-1}$ is also closed? I am I right?) however I didn't prove that $A$ is a closed operator, I just have a feeling it is closed.

3) I don't see how the compactness of the operator $A^{-1}$ follows from the Poincaré inequality and Rellich–Kondrachov.

We can prove the existence and boundedness of the inverse Laplacian using the Riesz representation theorem for Hilbert spaces. First, let us define the bilinear form $B[ \ , \ ]$ on $H_0^1(\Omega)$ as follows: $$B[u, v] = \int_{\Omega} \sum_i \partial_{x_i} u \partial_{x_i}v.$$

It is possible to prove a couple of inequalities:

• $|B[u,u]| \leq || u ||^2_{H_0^1(\Omega)}$

• $|| u ||_{H_0^1(\Omega)}^2 \leq c B[u, u]$ for a suitably chosen value of $c$.

The first inequality is obvious. The second isn't much harder - it just requires some fiddling around with the Poincare inequality.

These inequalities tell us that the norm associated to the inner product $B[ \ , \ ]$ is equivalent to the original Sobolev norm $|| . ||_{H_0(\Omega)}^1$. Therefore, since $H_0^1(\Omega)$ is complete with respect to the Sobolev norm, it must also be complete with respect to $B[\ , \ ].\$ As a consequence, we can legitimately apply the Riesz representation theorem in $H_0^1(\Omega)$ using $B[ \ , \ ]$ instead of $( \ , \ )_{H_0^1(\Omega)}$ as our inner product.

Now let's use the Riesz representation theorem in this way to deduce that the Laplacian operator has a bounded inverse. To be more precise, we want to show that, for every $g \in L^2(\Omega)$, there exists a unique $u_g \in H_0^1(\Omega)$ such that $$B[u_g, v ] = \int_{\Omega} g v \ \ \ \ \ \ \ \ \ {\rm for \ all \ } v \in H_0^1(\Omega)$$ and moreover, the mapping $$g \mapsto u_g$$ is a bounded linear map from $L^2(\Omega)$ to $H_0^1(\Omega)$.

This conclusion does indeed follow from the Riesz representation theorem, and the way to apply the Riesz representation theorem here is to think of $v \mapsto \int_\Omega g v$ as a linear functional on $H_0^1(\Omega)$ whose norm is no greater than $|| g ||_{L^2(\Omega)}$.

Notice that the $u_g$ that we have constructed is a solution to the equation $- \nabla^2 u = g$ in the weak sense. So if we are content to use the notation $\mathcal L$ for the $- \nabla^2$ operator, then we may as well use the notation $\mathcal L^{-1}$ for our newly constructed bounded operator $L^2(\Omega) \to H_0^1(\Omega)$ sending $g \mapsto u_g$.

Having defined our bounded inverse operator $\mathcal L^{-1} : L^2(\Omega) \to H_0^1(\Omega)$, I'll now discuss eigenfunctions. This is where we get to apply the Rellich-Kondrachov theorem and the spectral theorem for compact operators.

Let us define a weak eigenfunction of the Laplacian $\mathcal L$ (corresponding to the eigenvalue $k$) to be a $u \in H_0^1(\Omega)$ such that $$B[u, v] = k \int_\Omega u v \ \ \ \ \ \ \ \ \ {\rm for \ all \ } v \in H_0^1(\Omega)$$ We can immediately rephrase this definition in terms of our inverse operator $\mathcal L^{-1}$: A function $u \in H_0^1(\Omega)$ is a weak eigenfunction of $\mathcal L$ with eigenvalue $k$ if and only if $$u = k \left( \mathcal L^{-1} (u)\right),$$ Notice that the $u$ on the right-hand side of this equation is thought of as an element of $L^2(\Omega)$ whereas the $u$ on the left-hand side is thought of as an element of $H_0^1(\Omega)$. This makes sense, because $H_0^1(\Omega) \subset L^2(\Omega)$

I'm now going to massage this definition into a form that we can apply the spectral theorem to. Let's use the symbol $\iota$ to denote the inclusion $H_0^1(\Omega) \hookrightarrow L^2(\Omega)$. If you think about it, the previous paragraph can be written like this: A function $u \in L^2(\Omega)$ is a weak eigenfunction of $\mathcal L$ (and, in particular, is contained within the subspace $H_0^1(\Omega) \subset L^2(\Omega)$) iff it satisfies $$u = k \left( (\iota \circ \mathcal L^{-1}) (u)\right).$$

But $\iota : H_0^1(\Omega) \hookrightarrow L^2(\Omega)$ is a compact operator by Rellich, and $\mathcal L^{-1} : L^2(\Omega) \to H_0^1(\Omega)$ is a bounded operator, so the composition $$\iota \circ \mathcal L^{-1} : L^2(\Omega) \to L^2(\Omega)$$ is compact. The composition $\iota \circ \mathcal L^{-1}$ is also self-adjoint (and to check this, it suffices to verify self-adjointness on $C_{c}^\infty(\Omega)$, which is dense in $L^2(\Omega)$ and is contained inside $H_0^1(\Omega)$).

We can therefore legitimately apply the spectral theorem to $\iota \circ \mathcal L^{-1}$. This tells us that the weak eigenfunctions of $\mathcal L$ form a complete (countable) orthogonal basis for the orthogonal complement of the kernel of $\iota \circ \mathcal L^{-1}$ within $L^2(\Omega)$. But the kernel of $\iota \circ \mathcal L^{-1}$ is zero (since any $u$ in this kernel obeys $\int_\Omega uv = 0$ for every $v \in H_0^1(\Omega)$, and in particular, for every $v \in C_c^\infty(\Omega)$). The conclusion then is that the weak eigenfunctions of $\mathcal L$ form a complete (countable) orthogonal basis for $L^2(\Omega)$.

At the moment, these eigenfunctions are only weak eigenfunctions, living in $L^2(\Omega)$, and obeying only the weak condition $B[u, v] = k [u , v]$ for $v \in H_0^1(\Omega)$. It would be nice if we could show that these eigenfunctions are genuine smooth functions in $C^\infty(\Omega)$ obeying $\mathcal L u = k u$!

We can prove this as follows: By a regularity theorem in Evans Chapter 6.3, any weak solution $u$ to $\mathcal L u = f$, where $f$ is an element of $H^m(\Omega)$, must automatically be in $H_{\rm loc}^{m + 2}(\Omega)$. In our case, $u$ is a weak solution to $\mathcal L u = k u$. So the fact that $u$ is in $L^2(\Omega)$ implies that $u$ is in $H_{\rm loc}^2(\Omega)$, which in turn implies that $u$ is in $H_{\rm loc}^4(\Omega)$, which in turn implies that $u$ is in $H_{\rm loc}^6(\Omega)$, etc. Thus, $u$ is in $H_{\rm loc}^m(\Omega)$ for all $m$. But then, by the Sobolev inequalities, $u$ must also be in $C^\infty(\Omega)$, and we are done.

With further conditions on the smoothness of $\mathcal \Omega$, I believe it it is possible to prove that $u$ is also in $C^\infty_c(\bar \Omega)$, where $\bar \Omega$ is the closure of $\Omega$. (See Evans 6.3.) It then makes sense to evaluate $u$ on the boundary $\partial \Omega$, and the fact that $u$ is in $H_0^1(\Omega)$ rather than just $H^1(\Omega)$ ensures that $u|_{\partial \Omega} = 0$ (see Evans 5.5), which is to say that $u$ satisfies the Dirichlet boundary condition.

• Thanks, or, if I am not mistaken, we could claim from the beginning that the $u_g$ obtained by the Riesz representation theorem is in fact in $H^1_0 \cap H^2$ using the technical regularity apporachs in Evans. And thus we no longer have to deal with enigenfunctions in the weak sense. Aug 20, 2017 at 23:23
• Notice that $A$ is the Riesz isomorphism between $H_0^1$ and $H^{-1}$ as it is defined via $$\langle Au,v \rangle_{H^{-1} \times H_0^1} = (u,v)_{H_0^1}$$ and therefore its inverse exists and is continuous (hence bounded).

• We have $$L^2 \xrightarrow{\,J\,} H^{-1} \xrightarrow{A^{-1}} H_0^1 \xrightarrow{\text{Id}} L^2$$ where $J:L^2 \to H^{-1}$ such that $\langle Jh,v\rangle_{H^{-1} \times H_0^1}=(h,v)_{L^2}.$ And thus we consider the mapping $$\text{Id} \circ A^{-1} \circ J:L^2 \to L^2$$ which is mostly denoted just by $A^{-1}$. Notice that $J$ is the composition of the Riesz isomorphism of $L^2$ and the restriction mapping from $(L^2)' \to H^{-1}$. The mapping $\text{Id} \circ A^{-1} \circ J$ is compact as it is the composition of a compact operator ($\text{Id}$ by Rellich) and a continuous operator ($A^{-1}J$). Self-adjointness follows by Lax-Milgram's lemma since for all $f \in H^{-1}$ there is a unique $u \in H_0^1$ such that for all $v \in H_0^1$ $$(u,v)_{H_0^1}=(f,v)_{L^2}=\langle Jf,v \rangle_{H^{-1} \times H_0^1}=\langle AA^{-1}Jf,v \rangle_{H^{-1} \times H_0^1}=(A^{-1}Jf,v)_{H_0^1}$$ and taking the test function $v=A^{-1}Jg$ for a $g \in L^2$ yields $$(f,A^{-1} Jg)_{L^2}=(A^{-1}Jf,A^{-1} Jg)_{H_0^1}=\langle AA^{-1} Jg,A^{-1}Jf \rangle_{H^{-1} \times H_0^1}=(g,A^{-1}Jf)_{L^2}.$$ Finally, the spectral theorem yields an orthonormal basis in $L^2$.